Answer
$ \displaystyle \frac{(x-3)(x+3)}{ (x^{2}+1)^{1/2}}$
Work Step by Step
$(x^{2}+1)^{\frac{1}{2}}=(x^{2}+1)^{-\frac{1}{2}+1}=(x^{2}+1)^{-\frac{1}{2}}\cdot(x^{2}+1)$
factor out $(x^{2}+1)^{-1/2}$
$= (x^{2}+1)^{-1/2}\left[ (x^{2}+1)-10 \right]$
$=(x^{2}+1)^{-1/2}\left[ x^{2}-9 \right]$
the brackets hold a difference of squares, $x^{2}-3^{2}$,
use $a^{2}-b^{2}=(a+b)(a-b)$
= $(x^{2}+1)^{-1/2}(x-3)(x+3)\qquad $... apply $a^{-n}=\displaystyle \frac{1}{a^{n}}$
$= \displaystyle \frac{(x-3)(x+3)}{ (x^{2}+1)^{1/2}}$
